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I am trying to understand an identity of the $\delta$-function written on this Wikipedia page: \begin{equation} \int \mathrm{d} x \; f(x) \delta[g(x)] = \sum\limits_i \frac{f(x_i)}{\left| \frac{dg(x_i)}{dx}\right|} \tag{1} \end{equation} where $x_i$ are the zeros of $g(x)$ (i.e. $g(x_i)=0$). Now, my question is about the denominator on the right-hand side of equation $(1)$. Is that supposed to be a Jacobian determinant: \begin{equation} \left|\frac{dg(x_i)}{dx}\right| \overset{?}{=} \mathrm{det}\left(\frac{dg(x_i)}{dx}\right) \end{equation} or is it supposed to mean the modulus? The reason I think it might be a Jacobian is because Wiki mentions that we can use the following identity to change variables of integration: \begin{equation} \int_{\mathbf{R}} \delta\bigl(g(x)\bigr) f\bigl(g(x)\bigr) |g'(x)|\,dx = \int_{g(\mathbf{R})} \delta(u)f(u)\, du \end{equation}

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marked as duplicate by Chris Janjigian, vonbrand, LutzL, TMM, M Turgeon Mar 11 '14 at 1:07

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  • $\begingroup$ And has an answer there that is much more explicit than my answer. $\endgroup$ – LutzL Mar 11 '14 at 0:33
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The first variant. Everything is in one real variable, so you do not get Jacobian matrices to compute determinants.

The best way to understand that identity is to think of a delta-approximating sequence with compact support, for instance based on the quadratic or cubic B-Spline. Then consider small disjoint intervals around the roots of $g$, make the index in the delta-approximation so large that the support of the approximation is inside the images of these intervals, and perform standard parameter substitution on each of the intervals.

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  • $\begingroup$ Thanks for your answer. Just to be clear, if the integral was over multiple variables, and the functions $f$ and $g$ depended on those multiple variables, then we do compute the Jacobian matrices? $\endgroup$ – Hunter Mar 11 '14 at 1:27
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    $\begingroup$ Then the same applies, only that instead of intervals you take balls around the roots where $g'(x)$ is invertible. And then indeed it is the absolute value of the determinant in the denominator. $\endgroup$ – LutzL Mar 11 '14 at 1:45

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